Theorem sbalex 2278

Description: Equivalence of two ways to express proper substitution of a setvar for another setvar disjoint from it in a formula. This proof of their equivalence does not use df-sb 2099.

That both sides of the biconditional express proper substitution is proved by sb5 2309 and sb6 2121. The implication "to the left" is equs4v 2032 and does not require ax-10 2177 nor ax-12 2212. It also holds without disjoint variable condition if we allow more axioms (see equs4 2445). Theorem 6.2 of [Quine] p. 40. Theorem equs5 2489 replaces the disjoint variable condition with a distinctor antecedent. Theorem equs45f 2488 replaces the disjoint variable condition on 𝑥, 𝑡 with the nonfreeness hypothesis of 𝑡 in 𝜑. (Contributed by NM, 14-Apr-2008.) Revised to use equsexv 2302 in place of equsex 2447 in order to remove dependency on ax-13 2401. (Revised by BJ, 20-Dec-2020.) Revise to remove dependency on df-sb 2099. (Revised by BJ, 21-Sep-2024.) (Proof shortened by SN, 14-Aug-2025.)

Assertion

Ref	Expression
sbalex	⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝑡 → 𝜑))

Distinct variable group:   𝑥,𝑡

Allowed substitution hints:   𝜑(𝑥, 𝑡)

Proof of Theorem sbalex

Step	Hyp	Ref	Expression
1		nfe1 2186	. . 3 ⊢ Ⅎ𝑥∃𝑥(𝑥 = 𝑡 ∧ 𝜑)
2		ax12ev2 2215	. . 3 ⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) → (𝑥 = 𝑡 → 𝜑))
3	1, 2	alrimi 2249	. 2 ⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) → ∀𝑥(𝑥 = 𝑡 → 𝜑))
4		equs4v 2032	. 2 ⊢ (∀𝑥(𝑥 = 𝑡 → 𝜑) → ∃𝑥(𝑥 = 𝑡 ∧ 𝜑))
5	3, 4	impbii 212	1 ⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝑡 → 𝜑))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1567  ∃wex 1811

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-10 2177  ax-12 2212

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-nf 1816

This theorem is used by:  sb5  2309  dfsb7  2312  alexeqg  3604  regsfromsetind  37419  pm13.196a  45491